Material Removal Processes: “ Chip-forming”

Orthogonal cutting: the cutting edge is perpendicular to the cutting velocity, so the process can be treated in two dimensions. Symbols: $\alpha$ rake angle, $\phi$ shear angle, $\beta$ friction angle ($\tan\beta=\mu$), $t_0$ uncut chip thickness, $t_c$ chip thickness, $b$ width of cut.

Merchant circle

The resultant force $R$ in the chip is balanced by the equal and opposite force from the workpiece, and the circle with diameter $R$ contains all components.

 Chip      Tool
  \   shear plane (angle phi)
   \    /|
    \  / |  Rake angle alpha
 ----O----+-----> V (work moves right)

 Pairs of components of the same resultant R:
   (Fc , Ft)  - measured by dynamometer
   (Fs , Fn)  - along / normal to shear plane
   (F  , N )  - along / normal to tool face
 All lie on a circle of diameter R through O.

The simplified picture: $F_c$ and $F_t$ are measured by a dynamometer (horizontal and vertical); $F_s$ and $F_n$ act along and normal to the shear plane; $F$ and $N$ act along and normal to the tool face. All six components are the sides of right triangles with the common hypotenuse $R$.

Force relations

Resolving $F_c$ and $F_t$ along the shear plane (inclined at $\phi$ to the cutting direction):

$$F_s = F_c\cos\phi - F_t\sin\phi$$ $$F_n = F_c\sin\phi + F_t\cos\phi$$

Resolving along and normal to the tool face (rake angle $\alpha$):

$$F = F_c\sin\alpha + F_t\cos\alpha$$ $$N = F_c\cos\alpha - F_t\sin\alpha$$

Coefficient of friction: $\mu = F/N = \tan\beta$.

Chip thickness ratio and shear angle:

$$r = \frac{t_0}{t_c} = \frac{\sin\phi}{\cos(\phi-\alpha)}\quad\Rightarrow\quad \tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha}$$

Merchant's relation

Shear stress on the shear plane, whose area is $A_s = b\,t_0/\sin\phi$:

$$\tau = \frac{F_s}{A_s} = \frac{R\cos(\phi+\beta-\alpha)\,\sin\phi}{b\,t_0}$$

Since $F_c = R\cos(\beta-\alpha)$, eliminating $R$:

$$F_c = \frac{\tau\,b\,t_0\cos(\beta-\alpha)}{\sin\phi\,\cos(\phi+\beta-\alpha)}$$

Merchant assumed that the shear angle adjusts itself so that the cutting force (and so power) is a minimum, with $\tau$ constant. Setting $dF_c/d\phi = 0$ means the denominator $\sin\phi\cos(\phi+\beta-\alpha)$ is a maximum:

$$\cos\phi\cos(\phi+\beta-\alpha) - \sin\phi\sin(\phi+\beta-\alpha) = \cos(2\phi+\beta-\alpha) = 0$$ $$2\phi + \beta - \alpha = 90^\circ \quad\Rightarrow\quad \phi = 45^\circ + \frac{\alpha}{2} - \frac{\beta}{2}$$

Meaning

A larger rake angle or lower friction increases the shear angle, reduces the chip thickness and cutting force. Practical results agree only roughly with the formula; hence Lee and Shaffer's relation $\phi = 45^\circ + \alpha - \beta$ is sometimes used.

Given: $t_0 = 0.25$ mm, $t_c = 0.75$ mm, $b = 3$ mm, $\alpha = 10^\circ$, $F_c = 900$ N, $F_t = 450$ N, $V = 90$ m/min $= 1.5$ m/s.

Shear angle

Chip thickness ratio $r = t_0/t_c = 0.25/0.75 = 0.3333$.

$$\tan\phi = \frac{r\cos\alpha}{1 - r\sin\alpha} = \frac{0.3333\times0.9848}{1 - 0.3333\times0.1736} = \frac{0.3283}{0.9421} = 0.3485$$ $$\phi = 19.2^\circ$$

Forces on the tool face

$$F = F_c\sin\alpha + F_t\cos\alpha = 900(0.1736) + 450(0.9848) = 156.3 + 443.2 = 599.4\ \text{N}$$ $$N = F_c\cos\alpha - F_t\sin\alpha = 900(0.9848) - 450(0.1736) = 886.3 - 78.1 = 808.2\ \text{N}$$ $$\mu = \frac{F}{N} = \frac{599.4}{808.2} = 0.742,\qquad \beta = \tan^{-1}0.742 = 36.6^\circ$$

Shear force and shear stress

$$F_s = F_c\cos\phi - F_t\sin\phi = 900(0.9444) - 450(0.3289) = 850.0 - 148.0 = 702\ \text{N}$$

Shear plane area:

$$A_s = \frac{b\,t_0}{\sin\phi} = \frac{3\times0.25}{0.3289} = 2.28\ \text{mm}^2$$ $$\tau = \frac{F_s}{A_s} = \frac{702}{2.28} = 308\ \text{MPa}$$

Power and specific energy

$$P_c = F_c V = 900\times1.5 = 1350\ \text{W} = 1.35\ \text{kW}$$

Metal removal rate $= b\,t_0\,V = 3\times0.25\times1500 = 1125\ \text{mm}^3/\text{s}$.

$$u = \frac{P_c}{MRR} = \frac{1350}{1125} = 1.2\ \text{J/mm}^3$$

Answer: $\phi = 19.2^\circ$, $F = 599$ N, $N = 808$ N, $\mu = 0.74$, $F_s = 702$ N, $\tau = 308$ MPa, $P = 1.35$ kW, $u = 1.2$ J/mm$^3$.

Taylor's equation: the cutting speed $V$ and tool life $T$ (minutes to reach a given flank wear, usually 0.3 mm) are related by

$$V\,T^{\,n} = C$$

n depends mainly on the tool material (about 0.1-0.15 for HSS, 0.2-0.4 for carbide, 0.5-0.7 for ceramics), and C is the speed for a tool life of 1 minute.

Constants

Two tests give $V_1T_1^n = V_2T_2^n$:

$$\frac{V_1}{V_2} = \left(\frac{T_2}{T_1}\right)^{n}\;\Rightarrow\; \frac{100}{80} = \left(\frac{150}{60}\right)^{n}$$ $$n = \frac{\ln 1.25}{\ln 2.5} = \frac{0.2231}{0.9163} = 0.2435$$ $$C = V_1T_1^{n} = 100\times(60)^{0.2435} = 100\times2.710 = 271.0\ \text{m/min}$$

(Check: $80\times150^{0.2435} = 80\times3.388 = 271.0$.)

Speed for $T = 90$ min

$$V = \frac{C}{T^{n}} = \frac{271.0}{(90)^{0.2435}} = \frac{271.0}{2.992} = 90.6\ \text{m/min}$$

Tool life at 120 m/min

$$T = \left(\frac{C}{V}\right)^{1/n} = \left(\frac{271.0}{120}\right)^{4.107} = (2.258)^{4.107} = 28.4\ \text{min}$$

Comment on economics

A 20 percent rise in speed cuts the tool life from 60 to 28.4 min, i.e. by more than half. Machining time per piece falls, but the tool-change time and tool cost per piece rise. There is an economic cutting speed (minimum cost per piece) which lies below the speed of maximum production rate, and a speed of maximum profit between them. It is found by differentiating total cost (machining + tool-change + tool cost) with respect to speed.

Answer: $n = 0.2435$, $C = 271$ m/min, $V = 90.6$ m/min for 90 min life, $T = 28.4$ min at 120 m/min.

Desirable properties

1. Hot hardness - hardness retained at cutting temperature (up to 1000 degree C).
2. Toughness - resists shocks and interrupted cuts without chipping.
3. Wear resistance - long life against abrasion, adhesion and diffusion.
4. Chemical stability - does not react with the work material.
5. Low coefficient of friction and good thermal conductivity; and reasonable cost.

No material has all properties at once: hardness and toughness are in conflict.

Comparison

Cutting speeds rise in about the same order, from 20-60 m/min for HSS to more than 500 m/min for ceramics on cast iron.

Coated carbide

A thin layer (3-10 micrometre) of TiC, TiN, Al$_2$O$_3$ or a multilayer is deposited by CVD or PVD. The substrate provides toughness while the coating gives wear resistance, a thermal barrier and lower friction. Coated inserts allow 2-3 times the speed of uncoated ones, and give longer tool life.

Economics

Tool cost per piece depends on the tool price, number of regrinds or edges of an insert, and tool life. Disposable indexable inserts remove regrinding cost and give consistent size. A costly tool is justified when it reduces the cost per piece through higher speed, as in CNC and automatic production.

(a) Lathe operations and taper turning

The centre lathe holds the work between centres or in a chuck and the tool moves along or across the axis. Operations: plain (straight) turning, facing, step turning, chamfering, grooving, parting-off, taper turning, form turning, thread cutting, drilling, boring, reaming, knurling.

Methods of taper turning

1. Compound rest: the top slide is swivelled to half the taper angle and fed by hand. Suitable for short, steep tapers (angle up to about 45 degree).
2. Tailstock set-over: the tailstock is shifted sideways so that the work axis is inclined to the bed. Used for long gentle tapers; the centres are mismatched, which spoils the centre holes. Offset is given by $S = \frac{D-d}{2}\cdot\frac{L}{l}$ (L = bar length, l = length of taper).
3. Taper turning attachment: a guide bar at the set angle moves the cross slide automatically while the carriage travels. Gives accurate, long tapers without disturbing the centres.
4. Form tool: a tool shaped to the taper is plunged in; short tapers only.

(b) Numerical

Cutting speed

$$V = \frac{\pi D N}{1000} = \frac{\pi\times60\times500}{1000} = 94.2\ \text{m/min}$$

Material removal rate

$$MRR = f\,d\,V = 0.2\times 2.5\times 94.2\times1000 = 47\,124\ \text{mm}^3/\text{min} = 785\ \text{mm}^3/\text{s}$$

Power

$$P = u\times MRR = 2.5\times 785.4 = 1963\ \text{W} \approx 1.96\ \text{kW}$$

Machining time

$$T_m = \frac{L}{fN} = \frac{300}{0.2\times500} = 3.0\ \text{min}$$

Taper

Half angle for the compound rest:

$$\tan\theta = \frac{D-d}{2l} = \frac{60-40}{2\times200} = 0.05\;\Rightarrow\;\theta = 2.86^\circ$$

Tailstock offset for a 300 mm long bar:

$$S = \frac{D-d}{2}\cdot\frac{L}{l} = \frac{60-40}{2}\times\frac{300}{200} = 15\ \text{mm}$$

Answer: $V = 94.2$ m/min, $MRR = 47.1$ cm$^3$/min, $P = 1.96$ kW, time $= 3$ min; compound rest set at $2.86^\circ$; tailstock offset $= 15$ mm.

(a) Milling, shaper

Milling removes metal with a rotating multi-tooth cutter while the work is fed against it.

Shaper: a single-point tool is fixed on a reciprocating ram. The tool cuts on the forward stroke; on the return stroke it lifts (clapper box) and the table feeds the work sideways by a small amount. A quick-return mechanism (crank and slotted link) makes the return stroke faster. It is used for flat surfaces, slots and keyways.

(b) Drilling time

Length to be travelled:

$$L = t + 0.3D = 40 + 0.3\times20 = 46\ \text{mm}$$ $$T = \frac{L}{fN} = \frac{46}{0.15\times600} = 0.511\ \text{min} \approx 30.7\ \text{s}$$

Answer: drilling time $= 0.51$ min.